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{{redirect-distinguish|Variational theorem|variational principle}}
{{Refimprove|date=November 2011}}

In [[linear algebra]] and [[functional analysis]], the '''min-max theorem''', or '''variational theorem''', or '''Courant–Fischer–Weyl min-max principle''', is a result that gives a variational characterization of eigenvalues of [[Compact operator on Hilbert space|compact]] Hermitian operators on Hilbert spaces. It can be viewed as the starting point of many results of similar nature.

This article first discusses the finite dimensional case and its applications before considering compact operators on infinite dimensional Hilbert spaces. We will see that for compact operators, the proof of the main theorem uses essentially the same idea from the finite dimensional argument.

The min-max theorem can be extended to [[self adjoint operator]]s that are bounded below.

== Matrices ==

Let ''A'' be a ''n'' × ''n'' [[Hermitian matrix]]. As with many other variational results on eigenvalues, one considers the [[Rayleigh quotient|Rayleigh–Ritz quotient]] ''R''''A'': '''C'''''n'' \{0} → '''R''' defined by

:<math>R_A(x) = \frac{(Ax, x)}{(x,x)}</math>

where (·, ·) denotes the Euclidean inner product on '''C'''''n''. Clearly, the Rayleigh quotient of an eigenvector is its associated eigenvalue. Equivalently, the Rayleigh–Ritz quotient can be replaced by

:<math>f(x) = (Ax, x), \; \|x\| = 1.</math>

For Hermitian matrices, the range of the continuous function ''R''''A''(''x''), or ''f''(''x''), is a compact subset [''a'', ''b''] of the real line. The maximum ''b'' and the minimum ''a'' are the largest and smallest eigenvalue of ''A'', respectively. The min-max theorem is a refinement of this fact.

=== Min-max Theorem ===

Let ''A'' be a ''n'' × ''n'' [[Hermitian matrix]] with eigenvalues ''λ''1 ≥ ... ≥ ''λk'' ≥ ... ≥ ''λn'' then
:<math>
\lambda_k = \max \{ \min \{ R_A(x) \mid x \in U \text{ and } x \neq 0 \} \mid \dim(U)=k \}
</math>
and
:<math>
\lambda_k = \min \{ \max \{ R_A(x) \mid x \in U \text{ and } x \neq 0 \} \mid \dim(U)=n-k+1 \}
</math>
in particular,
:<math>
\lambda_n \leq R_A(x) \leq \lambda_1 \quad\forall x \in \mathbb{C}^n
</math>
and these bounds are attained when ''x'' is an eigenvector of the appropriate eigenvalues.

=== Proof ===

Since the matrix ''A'' is Hermitian it is diagonalizable and we can choose an orthonormal basis of eigenvectors {''u''1,...,''u''''n''} that is, ''u''''i'' is an eigenvector for the eigenvalue ''λ''''i'' and such that (''u''''i'', ''u''''i'' ) = 1 and (''u''''i'', ''u''''j'' ) = 0 for all ''i'' ≠ ''j''.

If ''U'' is a subspace of dimension ''k'' then its intersection with the subspace
:<math>
\text{span}\{ u_k, \ldots, u_n \}
</math>
isn't zero (by simply checking dimensions) and hence there exists a vector ''v≠0'' in this intersection that we can write as
:<math>
v = \sum_{i=k}^n \alpha_i u_i
</math>
and whose Rayleigh quotient is
:<math>
R_A(v) = \frac{\sum_{i=k}^n \lambda_i \alpha_i^2}{\sum_{i=k}^n \alpha_i^2} \leq \lambda_k
</math>
and hence
:<math>
\min \{ R_A(x) \mid x \in U \} \leq \lambda_k
</math>
And we can conclude that
:<math>
\max \{ \min \{ R_A(x) \mid x \in U \text{ and } x \neq 0 \} \mid \dim(U)=k \} \leq \lambda_k
</math>
And since that maximum value is achieved for
:<math>
U = \text{span}\{u_1,\ldots,u_k\}
</math>
We can conclude the equality.

In the case where ''U'' is a subspace of dimension ''n-k+1'', we proceed in a similar fashion: Consider the subspace of dimension ''k''
:<math>
\text{span}\{ u_1, \ldots, u_k \}
</math>
Its intersection with the subspace ''U'' isn't zero (by simply checking dimensions) and hence there exists a vector ''v'' in this intersection that we can write as
:<math>
v = \sum_{i=1}^k \alpha_i u_i
</math>
and whose Rayleigh quotient is
:<math>
R_A(v) = \frac{\sum_{i=1}^k \lambda_i \alpha_i^2}{\sum_{i=1}^k \alpha_i^2} \geq \lambda_k
</math>
and hence
:<math>
\max \{ R_A(x) \mid x \in U \} \geq \lambda_k
</math>
And we can conclude that
:<math>
\min \{ \max \{ R_A(x) \mid x \in U \text{ and } x \neq 0 \} \mid \dim(U)=n-k+1 \} \geq \lambda_k
</math>
And since that minimum value is achieved for
:<math>
U = \text{span}\{u_k,\ldots,u_n\}
</math>
We can conclude the equality.

== Applications ==

=== Min-max principle for singular values ===

The [[singular value]]s {''σk''} of a square matrix ''M'' are the square roots of eigenvalues of ''M*M'' (equivalently ''MM*''). An immediate consequence of the first equality from min-max theorem is

:<math>\sigma_k ^{\uparrow} = \min_{S_k} \max_{x \in S_k, \|x\| = 1} (M^* Mx, x)^{\frac{1}{2}}=
\min_{S_k} \max_{x \in S_k, \|x\| = 1} \| Mx \|.
</math>

Similarly,

:<math>\sigma_k ^{\downarrow} = \max_{S_k} \min_{x \in S_k, \|x\| = 1} \| Mx \|.</math>

=== Cauchy interlacing theorem ===

Let ''A'' be a symmetric ''n'' × ''n'' matrix. The ''m'' × ''m'' matrix ''B'', where ''m'' ≤ ''n'', is called a '''[[compression (functional analysis)|compression]]''' of ''A'' if there exists an orthogonal projection ''P'' onto a subspace of dimension ''m'' such that ''P*AP'' = ''B''. The Cauchy interlacing theorem states:

'''Theorem''' If the eigenvalues of ''A'' are ''α''1 ≤ ... ≤ ''αn'', and those of ''B'' are ''β''1 ≤ ... ''βj'' ... ≤ ''βm'', then for all ''j'' < m+1,

:<math>\alpha_j \leq \beta_j \leq \alpha_{n-m+j}.</math>

This can be proven using the min-max principle. Let ''βi'' have corresponding eigenvector ''bi'' and ''Sj'' be the ''j'' dimensional subspace ''Sj'' = span{''b''1...''bj''}, then

:<math>\beta_j = \max_{x \in S_j, \|x\| = 1} (Bx, x) = \max_{x \in S_j, \|x\| = 1} (P^*APx, x) \geq \min_{S_j} \max_{x \in S_j, \|x\| = 1} (Ax, x) = \alpha_j.</math>

According to first part of min-max,

:<math>\alpha_j \leq \beta_j.</math>

On the other hand, if we define ''S''''m''−''j''+1 = span{''b''''j''...''bm''}, then

:<math>\beta_j = \min_{x \in S_{m-j+1}, \|x\| = 1} (Bx, x) = \min_{x \in S_{m-j+1}, \|x\| = 1} (P^*APx, x)= \min_{x \in S_{m-j+1}, \|x\| = 1} (Ax, x) \leq \alpha_{n-m+j},</math>

where the last inequality is given by the second part of min-max.

Notice that, when ''n'' − ''m'' = 1, we have

:<math>\alpha_j \leq \beta_j \leq \alpha_{j+1}.</math>

Hence the name ''interlacing'' theorem.

== Compact operators ==

Let ''A'' be a [[Compact operator on Hilbert space|compact]], [[Hermitian]] operator on a Hilbert space ''H''. Recall that the [[spectrum (functional analysis)|spectrum]] of such an operator form a sequence of real numbers whose only possible [[cluster point]] is zero. Every nonzero number in the spectrum is an eigenvalue. It no longer makes sense here to list the positive eigenvalues in increasing order. Let the positive eigenvalues of ''A'' be

:<math>\cdots \le \lambda_k \le \cdots \le \lambda_1,</math>

where [[Multiplicity (mathematics)|multiplicity]] is taken into account as in the matrix case. When ''H'' is infinite dimensional, the above sequence of eigenvalues is necessarily infinite. We now apply the same reasoning as in the matrix case. Let ''Sk'' ⊂ ''H'' be a ''k'' dimensional subspace, and ''S' '' be the closure of the linear span ''S' '' = span{''uk'', ''u''''k'' + 1, ...}. The subspace ''S' '' has codimension ''k'' − 1. By the same dimension count argument as in the matrix case, ''S' '' ∩ ''Sk'' is non empty. So there exists ''x'' ∈ ''S' '' ∩ ''Sk'' with ||''x''|| = 1. Since it is an element of ''S' '', such an ''x'' necessarily satisfy

:<math>(Ax, x) \le \lambda_k.</math>

Therefore, for all ''Sk''

:<math>\inf_{x \in S_k, \|x\| = 1}(Ax,x) \le \lambda_k</math>

But ''A'' is compact, therefore the function ''f''(''x'') = (''Ax'', ''x'') is weakly continuous. Furthermore, any bounded set in ''H'' is weakly compact. This lets us replace the infimum by minimum:

:<math>\min_{x \in S_k, \|x\| = 1}(Ax,x) \le \lambda_k.</math>

So

:<math>\sup_{S_k} \min_{x \in S_k, \|x\| = 1}(Ax,x) \le \lambda_k.</math>

Because equality is achieved when ''Sk'' = span{''&u;''1...''&u;k''},

:<math>\max_{S_k} \min_{x \in S_k, \|x\| = 1}(Ax,x) = \lambda_k.</math>

This is the first part of min-max theorem for compact self-adjoint operators.

Analogously, consider now a ''k'' − 1 dimensional subspace ''S''''k''−1, whose the orthogonal compliment is denoted by ''S''''k''−1&perp;. If ''S' '' = span{''u''1...''uk''},

:<math>S' \cap S_{k-1}^{\perp} \ne {0}.</math>

So

:<math>\exists x \in S_{k-1}^{\perp} \, \|x\| = 1, (Ax, x) \ge \lambda_k.</math>

This implies

:<math>\max_{x \in S_{k-1}^{\perp}, \|x\| = 1} (Ax, x) \ge \lambda_k</math>

where the compactness of ''A'' was applied. Index the above by the collection of (''k'' − 1)-dimensional subspaces gives

:<math>\inf_{S_{k-1}} \max_{x \in S_{k-1}^{\perp}, \|x\|=1} (Ax, x) \ge \lambda_k.</math>

Pick ''S''''k''−1 = span{''u''1...''u''''k''−1} and we deduce

:<math>\min_{S_{k-1}} \max_{x \in S_{k-1}^{\perp}, \|x\|=1} (Ax, x) = \lambda_k.</math>

In summary,

'''Theorem (Min-Max)''' Let ''A'' be a compact, self-adjoint operator on a Hilbert space ''H'', whose positive eigenvalues are listed in decreasing order:

:<math>\cdots \le \lambda_k \le \cdots \le \lambda_1.</math>

Then

:<math>\max_{S_k} \min_{x \in S_k, \|x\| = 1}(Ax,x) = \lambda_k ^{\downarrow},</math>

:<math>\min_{S_{k-1}} \max_{x \in S_{k-1}^{\perp}, \|x\|=1} (Ax, x) = \lambda_k^{\downarrow}.</math>

A similar pair of equalities hold for negative eigenvalues.

== See also ==
* [[Courant minimax principle]]
* [[Max–min inequality]]

== References ==
*M. Reed and B. Simon, ''Methods of Modern Mathematical Physics IV: Analysis of Operators'', Academic Press, 1978.

[[Category:Functional analysis]]
[[Category:Articles containing proofs]]
[[Category:Theorems in functional analysis]]

Hypergeometric identity - Revision history

en>BD2412: minor fixes, mostly disambig links using AWB

en>RussBot: Robot: Editing intentional link to disambiguation page in hatnote per WP:INTDABLINK (explanation)